besiegefandomcom-20200222-history
Block Weight Problems
Notice: I have recently come back to this experiment (out of pure boredom) and looked over my calculations to see what amendments were needed. As well as rewriting this whole article to make it easier to understand. Aim When finding the perfect amount of lift for your machine, you will often find yourself using Balloons as oppose to Flying Blocks or Wings. It is to no ones surprise that the heavier the machine, the more Balloon Buoyancy you will need; how much Buoyancy do you exactly need? In this experimental write up I will explain the relationship between a Balloon's Buoyancy and the masses of Blocks in Besiege, I will even go over the already proposed masses of Blocks on this site and why they cannot be trusted. Quick Experiment The FANDOM page on Blocks' Masses states that a Wooden Block has the Mass of 0.50, the unit Mass is connected to the in-game adjustable Ballast Mass units. When connecting a Wooden block to a balloon with 0.00 string length, you'll find that 0.85 Balloon Buoyancy is needed to achieve lift. But when connecting a replica of that Balloon to a Ballast with the Mass of 0.50, and put them side by side, you'll see that the Wooden Block rises faster than the Ballast. Why? Key Terms Used - Balloon Buoyancy (BB) is the in-game unit for a Balloon's adjustable Buoyancy. - Ballast Mass (M) is the in-game unit for a Ballast's adjustable Mass, and is the basis of which we record and calculate all our Blocks. - 1:1 is the term used for when a Mass is increased by 0.01, so in turn the Balloon's Buoyancy is needed to increase by 0.01 respectively, to still achieve lift. - 4 and 5 are gaps of four and five 1:1 normal increases between an anomalous +0.02 increase in Balloon Buoyancy within a data set. Method Using a singular Balloon attached to nothing, slowly increase the BB from 0.20 by 0.01 and test each new BB until lift is achieved. Record the minimal BB of a singular Balloon, this will be important later. Place a Ballast above the ground and set it to 0.20 M, then attach a Balloon on top of the Ballast with the lowest string length and 0.20 BB. From there systematically increase the Buoyancy until the minimalist amount of BB is needed to lift the Ballast, record this BB alongside the M. Then increase the Ballast by 0.01 M, and find the new BB, record this, and repeat. This method should be repeated until a large data set is collected with noted patterns and/or anomalies. Results Singular Balloon = 0.26 BB * 0.21 M = 0.51 BB * 0.22 M = 0.52 BB * 0.23 M = 0.54 BB # * 0.24 M = 0.55 BB * 0.25 M = 0.56 BB * 0.26 M = 0.57 BB * 0.27 M = 0.58 BB * 0.28 M = 0.59 BB * 0.29 M = 0.61 BB # * 0.30 M = 0.62 BB * 0.31 M = 0.63 BB * 0.32 M = 0.64 BB * 0.33 M = 0.65 BB * 0.34 M = 0.67 BB # * 0.35 M = 0.68 BB * 0.36 M = 0.69 BB * 0.37 M = 0.70 BB * 0.38 M = 0.71 BB * 0.39 M = 0.72 BB * 0.40 M = 0.74 BB # * 0.41 M = 0.75 BB *Note: The results I have compiled for this experiment was larger than this, however for simplicity I have reduced it down to these easily reproducible values as they can still be used for further calculations and theory work. Discussion The reference Balloon was to simply see how much BB was needed to even lift up the Balloon itself, which came out to be 0.26 BB. This is useful for as any Mass we want to lift, we have to factor in 0.26 BB per Balloon. So to find the true Buoyancy needed to lift a Mass, take your current BB and subtract 0.26. A pattern soon arouse in my results, for small periods at a time, the BB to M was increasing 1:1 each time, increase by 0.01 M and you'll have to increase by 0.01 BB. However, at set moments the Buoyancy had to increase by 0.02 BB (noted by the #), even when the Mass was only increased by 0.01 M. This is evident by the 0.23 M having a 0.54 BB, when prior the 0.22 M only needing 0.52 BB. The +0.02 BB intervals can be categorised after a gap of four 1:1 increases (4), and a gap of five 1:1 increases (5), both coming one after another in the pattern of: 4, +0.02 BB, 5, +0.02 BB, 4... and so on. Reversing this pattern to find 0.00 M would give you a Buoyancy of 0.26 BB, the same Buoyancy given from our reference Balloon. With this in mind then, for a Mass of 0.20, the 0.5 BB lifting the Mass must be subtracted by 0.26 to yield 0.24 BB. So the true BB of a 0.20 M is 0.24 BB. I have also found when taking four 1:1 Masses from 0.25 M to 0.28 M (from a 5 gap) and placing them parallel to four 1:1 Masses from 0.30 M to 0.33 M (from a 4 gap), that the Masses in the 4 gap rise faster than the Masses in the 5 gap. Similarly, I tried this same method with two 4 gaps and two 5 and found that the heavier the Masses, the slower the gap gets. However a 4 gap will always be faster than its neighbouring 5 gaps. The point of noting the change in velocity can be used to theorise that eventually, with enough mass and buoyancy, that a true stabilised machine can be achieved. This can be worked out mathematically too, however the complexities of said maths deters me. Calculations To find a constant, k, to convert between BB and M you must divide the known true Balloon Buoyancy (BB - 0.26 * number of Balloons) by the known Mass. Example: (0.5 BB - 0.26*1) / 0.2 M = 1.2 k However if you do another example such as (0.97 BB - 0.26*1) / 0.6 M = 1.18333... k And once again and example of (0.61 BB - 0.26*1) / 0.29 M = 1.206896552 k The constant k changes, defeating the purpose of a constant. I found that the constant decreased when the weights increased (recorded from 0.2 M to 0.61 M), but for 0.62 M the constant increased dramatically, then proceeded to decreased again. So luckily I didn't have an exponential constant where I had to take natural logs. So what is the appropriate constant k? Well I took it upon myself to put my large data set of results into a Excel spreadsheet and use the formula =(SUM(All BB-0.26*n))/(SUM(All M) to find a reasonable value of 1.190584 as a constant k. With this constant, you'll still have to make a few assumptions on whether to round up or down the 0.001 decimal value. The Equation is then found to be M = (BBtot - 0.26 * n) / k Where, BBtot is the total Balloon Buoyancy given in-game, 0.26 is the required Balloon Buoyancy for a singular Balloon to lift itself, n is the number of Balloons, and k is the constant 1.190584. Likewise, BBtot = M * k + 0.26 * n Conclusion This is a big waste of my time. However! It comes with its advantages, this is a more reliable way of judging the Masses of Blocks and their required Buoyancy to create lift with a simple experimental set up and equation. These Masses aren't exact, they are based on Balloons lifting Blocks as apposed to achieving perfect aerial balance. If a perfect balance in the air was achieved, then I could base my calculations around that instead, to find the exact Masses of all Blocks. I propose to you, the reader (if you can be bothered) to either repeat my experiments to prove or disprove any of the statements or results I given, set up your own experiments and communicate to me what you did and your results to compare. And yes, I am aware of the article on Balloon Units, and I am able to confirm some values such as one Balloon having a Mass of 0.217, though through my calculation I got 0.218 M (0.26/k). And 1.183 being one of the conversions, however as I previously stated there are multiple constants for conversion and an average of all of them should be found for more reliability. I did my data set from 0.2 M to 0.63 M, I could have possibly missed out some key information that is further along, and with the 1:1 gaps getting slower as the weight increases, maybe perfect balance can be achieved. But for now, the Equation M=(BBtot-0.26*n)/k is the best I can do. If you have any questions or like to share your knowledge, feel free to ask, and I'll get to you in 4 months time. -MartynScience Category:Units of Measurement Category:Research